Question

Consider the reaction \(2 \mathrm{NO}_{2}(g) \longrightarrow \mathrm{N}_{2} \mathrm{O}_{4}(g) .\) (a) Using data from Appendix \(C\), calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). (b) Calculate \(\Delta G\) at \(298 \mathrm{~K}\) if the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{N}_{2} \mathrm{O}_{4}\) are \(0.40 \mathrm{~atm}\) and \(1.60 \mathrm{~atm}\), respectively.

Short answer

The standard Gibbs free energy change (ΔG°) for the reaction 2 NO₂(g) → N₂O₄(g) at 298 K is -4.71 kJ/mol. Using the given partial pressures of NO₂ = 0.40 atm and N₂O₄ = 1.60 atm, the Gibbs free energy change (ΔG) for the reaction at 298 K is approximately -1.82 kJ/mol.

Step-by-step solution

Write the balanced reaction equation

The balanced reaction equation is given in the exercise: 2 NO₂(g) → N₂O₄(g)

Calculate ΔG° at 298 K using Gibbs free energies of formation

We will use the given standard Gibbs free energies of formation in Appendix C to calculate ΔG° for the reaction at 298 K. We have: ΔG° = [G°(Product) * stoichiometric coefficient] - [G°(Reactant) * stoichiometric coefficient] Using the values for the standard Gibbs free energies of formation for the species (in kJ/mol): G°(NO₂) = 51.30 G°(N₂O₄) = 97.89 ΔG° = (1 * 97.89) - (2 * 51.30) ΔG° = 97.89 - 102.60 = -4.71 kJ/mol

Use the given partial pressures to find the reaction quotient, Q

We will find the reaction quotient, Q, using the given partial pressures: From the problem, we have: [NO₂] = 0.40 atm [N₂O₄] = 1.60 atm The reaction quotient Q is given by the expression: Q = \(\frac{[N_{2}O_{4}]}{[NO_{2}]^{2}}\) Now substitute the values: Q =\(\frac{1.60}{(0.40)^{2}}\) Q = 10

Calculate ΔG at 298 K using ΔG° and Q

We will use the equation relating ΔG, ΔG°, and Q: ΔG = ΔG° + RT ln(Q) Where R is the ideal gas constant 8.314 J/(mol * K) and T is the temperature (298 K in this case) Convert ΔG° to J/mol: ΔG° = -4.71 kJ/mol = -4710 J/mol ΔG = -4710 + (8.314 * 298 * ln(10)) ΔG ≈ -1824.87 J/mol ΔG ≈ -1.82 kJ/mol The Gibbs free energy change (ΔG) for the given reaction at 298 K with partial pressures of NO₂ = 0.40 atm and N₂O₄ = 1.60 atm is approximately -1.82 kJ/mol.

Key concepts

Chemical Equilibrium

Chemical equilibrium occurs in a reversible chemical reaction when the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, though they need not be equal.

For the reaction given in the exercise, \(2 \text{NO}_{2}(g) \rightleftharpoons \text{N}_{2}\text{O}_{4}(g)\), equilibrium would be reached when \(2 \text{NO}_{2}\) and \(\text{N}_{2}\text{O}_{4}\) are being produced and consumed at the same rate. The state of equilibrium is dynamic, meaning the molecules continue to react, but there is no net change in concentration of reactants and products. In any equilibrium process, the equilibrium position can be expressed quantitatively by the equilibrium constant (\(K_{eq}\)), which is closely related to the reaction quotient (\(Q\)) and Gibbs free energy, as discussed in the subsequent sections.

Reaction Quotient (Q)

The reaction quotient \(Q\) is a measure of the relative amounts of products and reactants present during a reaction at a given point in time. It is calculated in the same way as the equilibrium constant \(K_{eq}\) but is applicable at any point of the reaction, not just at equilibrium.

The reaction quotient is determined by the ratio of the concentrations (or partial pressures in the case of gases) of the products raised to the power of their coefficients in the balanced reaction, divided by the same for the reactants. For the gas phase reaction in our exercise, \(Q = \frac{{[N_{2}O_{4}]}}{{[NO_{2}]^{2}}}\). When \(Q = K_{eq}\), the system is at equilibrium. If \(Q < K_{eq}\), the reaction proceeds forward to form more products, and if \(Q > K_{eq}\), the reaction goes in the reverse direction to form more reactants.

Standard Gibbs Free Energy

The standard Gibbs free energy change (\(\text{Δ}G^{\text{°}}\)) for a reaction is the net change in free energy during a process when all reactants and products are in their standard states, which generally means 1 atm pressure for gases and 1 M concentration for solutions at a specified temperature, usually 25°C (298 K).

The \(\text{Δ}G^{\text{°}}\) for a reaction can be calculated from the standard free energies of formation (\(\text{Δ}G_{\text{f}}^{\text{°}}\)) of the products and reactants. The equation used in the exercise to calculate \(\text{Δ}G^{\text{°}}\) is \(\text{Δ}G^{\text{°}} = [\text{G}^{\text{°}}(\text{Product}) \times \text{stoichiometric coefficient}] - [\text{G}^{\text{°}}(\text{Reactant}) \times \text{stoichiometric coefficient}]\). This value is crucial for predicting whether a reaction will proceed spontaneously under standard conditions.

Thermodynamics in Chemistry

Thermodynamics in chemistry is the study of heat and energy transfer that accompanies chemical reactions and physical changes. The first and second laws of thermodynamics underpin the concepts of energy conservation and entropy, respectively, leading to the derivation of Gibbs free energy, \(G\), which predicts the spontaneity of a process.

Gibbs free energy combines the system's enthalpy (\(H\)), temperature (\(T\)), and entropy (\(S\)) in the equation \(G = H - TS\), indicating the maximum amount of work obtainable from a thermodynamic process at constant temperature and pressure. In the context of the exercise, the change in Gibbs free energy (\(\text{Δ}G\)) is used to determine the direction of the reaction. If \(\text{Δ}G < 0\), the process is spontaneous, whereas if \(\text{Δ}G > 0\), the process is non-spontaneous and requires external energy to proceed.